Educational Codeforces Round 33

A. Chess For Three Alex, Bob and Carl will soon participate in a team chess tournament. Since they are all in the same team, they have decided to practise really hard before the tournament. But it's a bit difficult for them because chess is a game fo

B. Beautiful Divisors Recently Luba learned about a special kind of numbers that she calls beautiful numbers. The number is called beautiful iff its binary representation consists of k + 1 consecutive ones, and then k consecutive zeroes. Some example

C. Rumor Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it

题目链接:Educational Codeforces Round 21 G. Anthem of Berland 题意: 给你两个字符串,第一个字符串包含问号,问号可以变成任意字符串. 问你第一个字符串最多包含多少个第二个字符串. 题解: 考虑dp[i][j],表示当前考虑到第一个串的第i位,已经匹配到第二个字符串的第j位. 这样的话复杂度为26*n*m*O(fail). fail可以用kmp进行预处理,将26个字母全部处理出来,这样复杂度就变成了26*n*m. 状态转移看代码(就是一个kmp

题目链接:Educational Codeforces Round 26 D. Round Subset 题意: 给你n个数,让你选其中的k个数,使得这k个数的乘积的末尾的0的个数最大. 题解: 显然,末尾乘积0的个数和因子2和因子5的个数有关. 然后考虑dp[i][j]表示选i个数,当前因子5的个数为j时,能得到因子2最多的为多少. 那么对于每个数,记录一下因子2和5的个数,做一些01背包就行了. 1 #include<bits/stdc++.h> 2 #define mst(a,b) me

题目链接:Educational Codeforces Round 23 F. MEX Queries 题意: 一共有n个操作. 1.  将[l,r]区间的数标记为1. 2.  将[l,r]区间的数标记为0. 3.  将[l,r]区间取反. 对每个操作,输出标记为0的最小正整数. 题解: hash后,用线段树xjb标记一下就行了. 1 #include<bits/stdc++.h> 2 #define ls l,m,rt<<1 3 #define rs m+1,r,rt<&l

题目链接:Educational Codeforces Round 23 D. Imbalanced Array 题意: 给你n个数,定义一个区间的不平衡因子为该区间最大值-最小值. 然后问你这n个数所有的区间的不平衡因子和 题解: 对每一个数算贡献,a[i]的贡献为 当a[i]为最大值时的 a[i]*(i-l+1)*(r-i+1) - 当a[i]为最小值时的a[i]*(i-l+1)*(r-i+1). 计算a[i]的l和r时,用单调栈维护.具体看代码,模拟一下就知道了. 然后把所有的贡献加起来.

题目链接:Educational Codeforces Round 25 F. String Compression 题意: 给你一个字符串,让你压缩,问压缩后最小的长度是多少. 压缩的形式为x(...)x(...)  x表示(...)这个出现的次数. 题解: 考虑dp[i]表示前i个字符压缩后的最小长度. 转移方程解释看代码,这里要用到kmp来找最小的循环节. 当然还有一种找循环节的方式就是预处理lcp,然后通过枚举循环节的方式. 这里我用的kmp找的循环节.复杂度严格n2. 1 #inclu

题目链接: Educational Codeforces Round 23 E. Choosing The Commander 题意: 一共有n个操作. 1.  插入一个数p 2.  删除一个数p 3.  询问有多少个数 使得 x^p<l 题解: 对于前两种操作用01trie就能解决. 对于对三个操作,我们考虑在trie上搜索. 1.  当l的bit位是1时,那边bit位是p的字数全部的数都会小于l,(因为p^p=0) 2.  当l的bit为是0时,那边只能向bit位是p的子树中搜. 这样算下来